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bio website justindomingue.com
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visits member for 2 years, 9 months
seen Dec 9 at 14:26

Dec
4
awarded  Notable Question
Dec
2
awarded  Popular Question
Nov
22
awarded  Popular Question
Sep
24
awarded  Popular Question
Jul
2
awarded  Curious
Apr
15
awarded  Popular Question
Jan
29
revised taylor series for cosx around 0
LAtex formatting
Jan
29
suggested approved edit on taylor series for cosx around 0
Jan
28
comment Proof that $AI_n = A$ using $Ab_i$.
Thanks for the clarification! I had a hard time visualizing the matrices but you made it very clear. Thanks again.
Jan
28
accepted Proof that $AI_n = A$ using $Ab_i$.
Jan
27
comment Proof that $AI_n = A$ using $Ab_i$.
OK so if I understand well, $I_n = [b_1 b_2 b_3 \dots b_n]$ where each $b_i$ is a column vector. When you say $A(b_1\dots b_n)$, is $(b_1\dots b_n)$ a matrix or $b_1 \cdot b_2 \cdot \dots \cdot b_n$?
Jan
27
comment Proof that $AI_n = A$ using $Ab_i$.
@ChristopherErnst, how is $I_n = (b_1\dots b_n)$? Is the definition using matrix multiplication?
Jan
27
asked Proof that $AI_n = A$ using $Ab_i$.
Nov
19
accepted Monotonic uniformly continuous function - Unique $f(t) = t$
Nov
18
comment Monotonic uniformly continuous function - Unique $f(t) = t$
OK, I understand your hint now. (it seems I was not in the right track... thanks!)
Nov
18
comment Monotonic uniformly continuous function - Unique $f(t) = t$
@BettyMock, I am actually using Bolzano's Intermediate Value Theorem which I think does not require the endpoints to be of different signs
Nov
18
comment Monotonic uniformly continuous function - Unique $f(t) = t$
@StephenMontgomery-Smith, I don't understand your hint. You want me to feed them into $|f(x)-f(x')| < |x-x'|$? Is the goal to show monotonicity or $f(t)=t$?
Nov
18
asked Monotonic uniformly continuous function - Unique $f(t) = t$
Nov
5
accepted Example to $\lim f(x)g(x)$ may not exist
Nov
5
comment Example to $\lim f(x)g(x)$ may not exist
That's actually what I had at first (with 1 instead of 5). If I take $g=1$, does $fg = f$ (even if $f$ is defined by part)